FREE NEC Code Questions and Answers

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A 45 KVA three phase transformer with a 115/230 delta configuration. What is the transformer secondary's high-leg voltage?

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A delta connected transformer high leg voltage = ( Line voltage /2 ) * 1/732 High leg voltage = 230/2 * 1.732 High leg voltage = 199 Volts

In a 4 in. * 4 in. wireway, how many 500 kcmil THWN conductors can be installed?

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Wireway area = 4 * 4 = 16 sq. in. Refer NEC [376.22(A)] The conductor fill area should be maximum 20% of wireway area. WireWay conductor fill = 16 * 0.2 = 3.2 sq. in. Area of 500 kcmil THHN = 0.7073 sq in. [Chapter 9, Table 5] Number of 500 kcmil THWN conductors = 3.2 / 0.7073 = 4.5 ( 4 conductors)

Under a pool or within _____ horizontally of the inside wall of a pool, underground wiring is not allowed:

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The National Electrical Code (NEC) in the United States does not permit underground wiring under a pool or within 5 feet horizontally from the inside wall of a pool.

What is the needed conductor size for GEC if a service is fed by three parallel sets of 400 kcmil conductors?

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The total underground conductor size = 3 * 400 = 1200 kcmil .Refer NEC Table 250.166 .The required GEC is 3/0 AWG for 1200 kcmil size

A pool, spa, hot tub, fountain, or the current tidal high water mark should not be located within _____ horizontally of audio system equipment fed by a branch circuit:

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According to NEC Article 680.22, audio system equipment supplied by a branch circuit must not be located within 5 feet horizontally of the inside wall of a pool, spa, hot tub, fountain, or within 5 feet horizontally of the prevailing tidal high water mark.

For a 4in * 4 in wireway, what is the maximum permitted conductor fill?

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Wireway Area = 4 in. * 4 in. = 16 sq. in. Refer : NEC 376.22(A) The conductor fill area should be maximum 20 % of the total wireway area. Wireway conductor fill = 16 * 0.2 = 3.2 sq. in.

For a constant load of 22 amperes at 750C, what size conductor should be used to protect against overcurrent?

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The size of branch circuit conductor to protect against over current must be more than 125 % of continuous load. 22 * 1.25 = 27.5 amperes. Refer : Table 310.15(B)(16). No 10 THWN with 35 amperes at 75 0C is suitable

A set of 450 kcmil feeder conductors is protected by a 400 breaker. With a 25 A over current device, a feeder fed from a 450 kcmil feeder disconnects. What is the smallest conductor size for these taps?

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Answer : 8 AWG The over current device 25 A is less than 10 percent of the rating of the feeder breaker (400A.) As per code it cannot be less than 40 A. Refer : NEC Table 310.15(B)(16) 8 AWG at 40A 60 C is the minimum size required.

If the overcurrent device rating is not specified, the overcurrent protection device of non-motor equipment shall not be greater than :

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The National Electrical Code (NEC) in the United States specifies that the overcurrent protection device for non-motor appliances must not exceed 20 amperes if the overcurrent device rating is not marked on the appliance.

Continuous load is defined as a load when the maximum current is anticipated to last for __ hours or more:

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According to the NEC, continuous loads are those where the maximum current is expected to last for three hours or longer at full rated load. Examples of continuous loads include heating equipment, lighting fixtures, and certain motors that run continuously for extended periods.

What is the general lighting load of a hospital with an area of 80,000 sq ft after the demand factor is applied?

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As per NEC Table 220.12 , The unit general lighting load for hospital is 2 VA per sq ft. So total lighting load of hospital = 80,000 * 2 = 1,60,000 VA As per NEC Table 220.42 The lighting load demand factors are as follows First 50000 or less 40% i.e 50,000 * 0.4 = 20000 Reaming is 20% 1,10,000 * 20 = 22000 So total lighting load after demand factor is applied = 20000 + 22000 = 42000 VA

For a 4000 W, 240 electric water heater, what size branch circuit conductor is required?

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Branch circuit conductor size should 125% of continuous load [ NEC 422.10(A)] Electric water heater current = 4000 / 240 = 16.6 Amperes Branch conductor size = 16.6 * 1.25 = 20.75 amperes As per branch conductor next size up rule 10 AWG with rating of 30 A at 60 C is suitable [ Refer NEC Table 310.15(B)(16)]

For a dwelling unit with a calculated load of 240 A, what size feeder conductors are required? The feeder conductors carry the whole load of the dwelling unit, and the service disconnect is rated for 250 A?

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Answer: 4/0 AWG Note: Feeder conductors that supply entire load of dwelling unit can have the rating of 83% of service disconnect of dwelling unit. The service conductor ampacity for 250 A service disconnect = 0.83 * 250 = 207.5 amperes Refer : NEC Table 310.15(B)(16) 4/0 AWG with ampacity 250 amperes ( 75 C) is suitable as service conductor

Type AC cables need to be supported and fastened at spaced no more than :

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Type AC (Armored Cable) cables, also known as "BX" cables, must be supported and secured at intervals not exceeding 4.5 feet, according to the National Electrical Code (NEC) in the United States.

For 12 AWG copper conductor, the maximum overcurrent protection is:

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The maximum overcurrent protection for a 12 AWG (American Wire Gauge) copper conductor is typically 20 Amperes. This means that the circuit breaker or fuse protecting the 12 AWG copper conductor should not have a rating higher than 20 Amperes.

The wires connecting the final branch circuit overcurrent device to the service equipment, an independently derived system, or another power source

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Feeder conductors are responsible for carrying electrical power from the source (service equipment, transformer, etc.) to various distribution points within a building or facility. These distribution points can be panelboards or other load centers where the final branch circuit overcurrent devices (circuit breakers or fuses) are located.